Problem 68 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\vec{a}, \vec{b}\), and \(\overrightarrow{\mathrm{c}}\) are unit vectors such that \(\vec{a}+2 \vec{b}+2 \overrightarrow{\mathbf{c}}=\overrightarrow{0}\), then \(|\vec{a} \times \overrightarrow{\mathrm{c}}|\) is equal to [Online April 15, 2018] (a) \(\frac{1}{4}\) (b) \(\frac{\sqrt{15}}{4}\) (c) \(\frac{15}{16}\) (d) \(\frac{\sqrt{15}}{16}\)

Step-by-Step Solution

Answer

\(|\vec{a} \times \overrightarrow{\mathrm{c}}|\) is \(\frac{\sqrt{15}}{4}\).

Step 1: Substitute and Rearrange

Given the vector equation \( \vec{a} + 2\vec{b} + 2\overrightarrow{\mathrm{c}} = \overrightarrow{0} \), we can rearrange to express \( \vec{a} \) in terms of \( \vec{b} \) and \( \overrightarrow{\mathrm{c}} \). Thus, \( \vec{a} = -2(\vec{b} + \overrightarrow{\mathrm{c}}) \).

Step 2: Explore Magnitude Constraints

Since \( \vec{a}, \vec{b} \), and \( \overrightarrow{\mathrm{c}} \) are unit vectors, they have magnitudes of 1. Therefore, \( |\vec{a}| = 1 \), while from the expression \( \vec{a} = -2(\vec{b} + \overrightarrow{\mathrm{c}}) \), it follows that \( |\vec{a}| = 2 |\vec{b} + \overrightarrow{\mathrm{c}}| \). Equating the magnitudes gives us, \( 2 |\vec{b} + \overrightarrow{\mathrm{c}}| = 1 \), hence, \( |\vec{b} + \overrightarrow{\mathrm{c}}| = \frac{1}{2} \).

Step 3: Use Vector Identity for Cross Product

Apply the identity \( \vec{a} \times \overrightarrow{\mathrm{c}} = 2(\vec{b} + \overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{c}} \). Using properties of cross products, \( \vec{b} \times \overrightarrow{\mathrm{c}} \) simplifies the expression to \( 2(\vec{b} \times \overrightarrow{\mathrm{c}}) \). This leads us to the need to find \( |\vec{b} \times \overrightarrow{\mathrm{c}}| \).

Step 4: Evaluate \(|\vec{b} \cdot \overrightarrow{\mathrm{c}}|^2\)

Using the identity \( |\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2(\vec{u} \cdot \vec{v}) \), where \( |\vec{u}| = |\vec{v}| = 1 \), apply it to \( |\vec{b} + \overrightarrow{\mathrm{c}}| = \frac{1}{2} \). Thus, \( \frac{1}{4} = 1 + 1 + 2(\vec{b} \cdot \overrightarrow{\mathrm{c}}). \) Solve for \( \vec{b} \cdot \overrightarrow{\mathrm{c}} \), leading to \( -\frac{7}{8} \).

Step 5: Determine \(|\vec{b} \times \overrightarrow{\mathrm{c}}|\)

The expression for \(|\vec{b} \times \overrightarrow{\mathrm{c}}|\) uses the Pythagorean identity, \(|\vec{b} \times \overrightarrow{\mathrm{c}}|^2 + (\vec{b} \cdot \overrightarrow{\mathrm{c}})^2 = 1.\) Substituting \( \vec{b} \cdot \overrightarrow{\mathrm{c}} = -\frac{7}{8} \), compute \( |\vec{b} \times \overrightarrow{\mathrm{c}}| = \sqrt{\frac{15}{64}} = \frac{\sqrt{15}}{8}. \)

Step 6: Final Calculation for \(|\vec{a} \times \overrightarrow{\mathrm{c}}|\)

Since \( \vec{a} \times \overrightarrow{\mathrm{c}} = 2(\vec{b} \times \overrightarrow{\mathrm{c}}) \), we can compute \(|\vec{a} \times \overrightarrow{\mathrm{c}}| = 2 \times \frac{\sqrt{15}}{8} = \frac{\sqrt{15}}{4}.\)

Key Concepts

Unit Vectors

Unit vectors play a crucial role in vector mathematics, serving as building blocks in various calculations. A unit vector is a vector that has a magnitude of 1. It essentially represents direction alone, without scaling the vector. In terms of notation, if we have a unit vector \( \vec{a} \), it is often expressed in terms of its components, for example, \( \vec{i}, \vec{j}, \vec{k} \) in a three-dimensional Cartesian coordinate system.

These unit vectors are pivotal when dealing with vector operations like the cross product, as they allow for simplifying complex vector calculations and help in defining directions in space with precision. When multiple unit vectors are involved, as in our problem, it is important to remember each vector's magnitude remains 1, even when combined with others.

Magnitude of a Vector

The magnitude of a vector is essentially the vector's length and is a fundamental concept when performing vector operations. Mathematically, the magnitude of a vector \( \vec{v} \) is denoted as \( |\vec{v}| \) and calculated using the formula: \[ \sqrt{v_{x}^2 + v_{y}^2 + v_{z}^2} \]where \( v_x, v_y, \) and \( v_z \) are the vector's components.

In problems involving unit vectors, the importance of magnitude is underscored by the fact that these vectors are defined with a magnitude of 1. However, when unit vectors are combined or multiplied by a scalar, as seen in the given exercise, the resultant vector's magnitude may change. Understanding how to find this new magnitude is essential for further operations like verifying vector identities or performing cross products.

Cross Product Properties

The cross product of two vectors results in a new vector that is perpendicular to the plane formed by the original vectors. When computing the cross product \( \vec{a} \times \vec{b} \), the magnitude of the resulting vector is given by the formula:\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta) \]where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). This operation is only defined in three-dimensional space.

The cross product has several key properties, such as being anti-commutative, meaning \( \vec{a} \times \vec{b} = - (\vec{b} \times \vec{a}) \), and distributive over vector addition, as in \( \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \). Understanding these properties assists in simplifying problems and calculating cross products efficiently, as demonstrated in the step-by-step solution.

Vector Identities

Vector identities are mathematical truths that simplify and relate different vector operations. In vector algebra, numerous identities make complex calculations more manageable, often appearing in exercises involving cross product or dot product.

One such identity used in the problem was the expression:\[ |\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2(\vec{u} \cdot \vec{v}) \]This identity connects the magnitude of a vector sum with the magnitudes and dot product of the individual vectors.

• Dot product: This involves conceptually understanding how the contribution of one vector in the direction of another can impact such identities.
• Cross product: Pythagorean identity \(|\vec{b} \times \overrightarrow{\mathrm{c}}|^2 + (\vec{b} \cdot \overrightarrow{\mathrm{c}})^2 = 1\) was used to find the cross product's magnitude, illustrating how these identities offer a consistent framework for solving vector problems.

Mastering these identities not only simplifies calculations but also provides deeper insights into the fundamental attributes of vector operations.